Why isn't tangential acceleration just $a$?

Question

If the tangential acceleration is $\mathrm d|v|/\mathrm dt$ then isn't it just the magnitude of the acceleration of the object because $\mathrm dv/\mathrm dt$ is acceleration?

Answer 1

You've got to be careful with notation when differentiating vectors, because it's very easy to make mistakes.

The acceleration is a vector and is a derivative of the velocity vector \begin{equation} \vec{a} = \frac{d \vec{v}}{d t} \end{equation} It is often useful to express this in terms of components. Let's stick with Cartesian coordinates, since then we don't have to worry about derivatives of the basis vectors. If we define $\hat{e}_i$ (where $i=x, y, z$) to be a set of three unit vectors in the $x, y, z$ directions, then since $d\hat{e}_i/dt$=0, we have that the components of the acceleration are related to the components of the time derivatives of the velocity vector by \begin{equation} a_i = \hat{e}_i \cdot \vec{a} = \hat{e}_i \cdot \frac{d \vec{v}}{d t} = \frac{d}{dt} \left(\hat{e}_i \cdot \vec{v}\right) = \frac{d v_i}{dt} \end{equation} where $\vec{A}\cdot \vec{B}$ represents the dot product between $\vec{A}$ and $\vec{B}$. The magntitude of the acceleration vector is \begin{equation} |\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} = \sqrt{\vec{a} \cdot \vec{a}} \end{equation}

Now let us differentiate the magnitude of the velocity vector. The magnitude is given by \begin{equation} |\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2} \end{equation} so \begin{eqnarray} \frac{d|\vec{v}|}{dt} &=& \frac{1}{2\sqrt{v_x^2 + v_y^2 + v_z^2}} \left(2 v_x \frac{dv_x}{dt} + 2 v_y \frac{dv_y}{dt} + 2 v_z \frac{dv_z}{dt}\right) \\ &=& \frac{v_x a_x + v_y a_y + v_z a_z}{\sqrt{v_x^2 + v_y^2 + v_z^2}} \\ &=& \frac{\vec{v} \cdot \vec{a}}{|\vec{v}|} \end{eqnarray}

Comparing the expressions we've derived for $|\vec{a}|$ and $\frac{d|\vec{v}|}{dt}$, it's clear that they are not equal. For one thing, $|\vec{a}$| is always positive, while $\frac{d|\vec{v}|}{dt}$ can be positive or negative since $|\vec{v}|$ can be growing or shrinking. For another thing, the magnitudes are also different in general. To give a concrete example, if $a_x=a$, $a_y=a_z=0$, $v_x=v_y=v/\sqrt{2}$, and $v_z=0$, then $|\vec{a}|=|a|$ and $\frac{d|\vec{v}|}{dt} = a / \sqrt{2}$.

Answer 2

The acceleration vector has two components.

• The tangential component (tangent to the path) equals the rate of change of speed $a_t = \dot{v}$. Speed is the magnitude of the velocity vector $v = \| \vec{v} \|$ and the velocity vector defines the tangent direction $\hat{e} = \vec{v} / \| \vec{v} \\$

• The normal component (perpendicular to the path) equals the acceleration due to rotation about a point $a_n = \frac{v^2}{r}$ where $r$ is the radius of curvature.

Together you have

$$ \boxed{ \vec{a} = \dot{v} \hat{e} + \frac{v^2}{r} \hat{n} }$$

You can always decompose any acceleration vector into the above components.